PROBLEM SET 1.2. KINETICS AND DIFFUSION

ANSWER KEY

1. A. The empirical formula of glucose is C6H12O6. What is its molecular weight?

C6H12O6 should have a molecular weight of (6 x 12 ) + (12 x 1) + (6 x 16) = 72 + 12 + 96 =

180 g mol-1

B. Isotonic glucose is 5% glucose w:v. How much glucose would we need to make 100 mL ofisotonic glucose?

5 g% means 5 g per dL; so we need 5 g of glucose to make up this 100mL of isotonic glucose.

2. A. You need to make 250 mL of a stock solution of 0.1 M Na2 ATP. Its formula weight is 605.2g mol-1. How much Na2 ATP should you weigh out?

Amount of ATP = C x V = 0.1 mol L-1 x 0.25 L = 0.025 mol

0.025 mol x 605.2 g mol-1 = 15.13 g

B. Your advisor is skeptical of your abilities. He wants you to check out the 0.1 M ATP solutionand tells you to do it spectrophotometrically. Spectrophotometry relies on the differentabilities of chemicals to absorb light of specific wavelengths. A diagram of aspectrophotometer is shown below.

Fig. 1.PS2.1. Light path in a single-beamspectrophotometer. The view is from above. Light from a source is collimited (making anarrow beam) and passed through amonochromator that selects a narrow band ofwavelength of light to be passed through thesample. A photomultiplier tube (PMT) detectsthe light and measures its intensity. Comparisonof this intensity, I, to the intensity when thesample is missing, I0, allows calculation of theabsorbance. Absorbance is recorded with timeor as a function of wavelength.

At particular wavelengths, chemicals absorb light according to their chemical structure and theirconcentration. The law governing the absorption of light is the Beer-Lambert Law:

A = å C d

where A is the absorbance, å is a constant that depends on the chemical and typically varies with thewavelength of light - it is the molar extinction coefficient and is in units of M-1 ; C is the concentration of the

1.PS2.1

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chemical (in M) and d is the pathlength. The molar extinction coefficient is defined for a pathlength of 1 cm. The absorbance is defined as

A = log ( I0 / I )

where I0 is the incident light intensity and I is the transmitted light intensity.

Your advisor tells you that å259 = 15.4 x 103 M-1; this is the molar extinction coefficient of ATP at awavelength of incident light of 259 nm. He tells you to make a dilution of the stock by taking 25 ìL of the stocksolution and diluting it to 100 mL. What absorbance do you expect of the final diluted solution, if you madeit up correctly, at ë = 259 nm?

Dilution gives C1 x V1 = C2 x V2 ; 0.025 mL x 0.1 M = 100 mL x CATP

CATP = 0.025 mL x 0.1 M / 100 mL = 25 x 10-6 M

A = å C l = 15.4 x 103 M -1 x 25 x 10-6 M = 0.385

3. A. The molecular weight of ryanodine is 493.54 g mol-1. You want to make 10 mL of a 10 mMstock solution. How much ryanodine should you weigh out?

10 mL = 0.010 L; 0.010 L x 10 x 10-3 mol L-1 = 0.1 x 10-3 mol;

1 x 10-4 mol x 493.54 g mol-1 = 0.04935 g = 49.35 mg

B. You make a dilution of the 10 mM ryanodine stock by pipetting 10 ìL of the stock solution toa 10 mL volumetric flask and adding water to the mark. You measure the absorbance as afunction of wavelength (against a water blank) and find a peak at 271 nm with an absorbanceof 0.179. What is å271 for ryanodine? (See 1.PS2 problem #2 for a discussion of the Beer-Lampert Law and a definition of the molar extinction coefficient).

Dilution gives C1 x V1 = C2 x V2 ; 0.01 mL x 10 mM = 10 mL x CRY

CRY = 0.01 mL x 10 mM / 10 mL = 0.01 mM = 10 x 10-6 M

A = å C l ; 0.179 = å x 10 x 10-6 M ; å = 0.179 /10 x 10-6 M = 1.79 x 104 M-1

4. A. Magnesium chloride has a formula of MgCl2 C 6H2O. What is its formula weight?

Mg has an atomic weight of 24.31 g mol-1; Cl has an atomic weight of 35.45 g mol-1. H2O hasa formula weight of 18 g mol-1. Thus the formula weight of MgCl2 C6 H2O is 24.31 + 2 x35.45 + 6 x 18 = 203.21 g mol-1.

B. You desire to make 1 L of 0.1 MgCl2 solution. How much MgCl2 C 6H2O should you weighout?

Amount = volume x concentration = 1 L x 0.1 M = 0.1 mol; 0.1 mol x 203.21 g mol-1 =20.32 g

C. You need to make 25 mL of a 25 mM solution of MgCl2. How much of the 0.1 M stocksolution do you add to the 25 mL volumetric flask?

Dilution gives C1 x V1 = C2 x V2 ; 25 x 10-3 L x 25 x 10-3 M = X mL x 0.1 M ;

1.PS2.2

X = 6250 x 10-6 L = 6.25 mL

5. The extracellular fluid volume varies with the size of the person. Suppose in an individual wedetermine that the ECF is 14 L. The average [Na+] in the ECF is about 143 mM.

A. What is the total amount of Na+ in the ECF, in moles? In grams?

Amount = Volume x concentration; = 14 L x 143 x 10-3 mol L-1 = 2002 x 10-3 mol = ; 2.002 mol.

We convert this to grams by multiplying by the gram atomic weight of Na = 22.99 g mol-1: 2.002 mol x 22.99 g mol-1 = 46.03 g

B. Suppose this person works out and sweats 1.5 L with an average [Na+] of 50 mM. Duringthis time the urine output is 30 mL with an average [Na+] of 600 mM. How much Na+ is lostduring the workout?

1.5 L x 50 x 10-3 mol L-1 = 0.075 mol = 1.72 g Na+ lost in the sweat0.03 L x 600 x 10-3 mol L-1 = 0.018 mol = 0.41 g Na+ lost in the urineTotal loss = 2.13 g Na

C. If the person does not drink fluids at all during the workout, what will be the [Na+] in theplasma at the end of the workout? Assume that all of the fluid in the sweat and urineoriginated from the ECF.

Final ECF = 14 L - 1.53 = 12.47 L; Final Na+ = 2.002 mol - 0.075 - 0.018 = 1.909 molFinal [Na+] = 1.909 mol/ 12.47 L = 0.153 M

6. The body normally produces about 2 g of creatinine per day. The amount varies with individuals andis approximately proportional to the muscle mass. It is excreted through the kidneys according toUrinary excretion of creatinine = GFR x Plasma concentration of creatinine. If the GFR is 120 mLmin-1, what is the plasma concentration of creatinine at steady-state? Hint: assume the body is atsteady-state with respect to creatinine.

At steady state, the rate of excretion of creatinine has to match its rate of production. The rate ofproduction is given at 2 g/day. The urinary excretion is given as GFR x P Creatinine. The GFR is givenas 120 mL min-1. Thus we have

2 g day-1 = 120 mL min-1 x 60 min hr-1 x 24 hr day-1 x P Creatinine;

P Creatinine = 2 g day-1 /172800 mL day-1 = 1.157 x 10-5 g mL-1 = 1.157 mg%

7. Just before noon, your plasma glucose concentration was 100 mg dL-1. This plasma glucose isapproximately evenly distributed among 3.5 L of plasma and 10.5 L of interstitial fluid that comprisesyour 14 L of ECF. Glucose is readily distributed in both compartments. You drink a can of soda thatcontains 35 g of glucose.

A. How much would you blood glucose rise if all the glucose in the soda was absorbed and noneof it was metabolized?

The 35 g of glucose would be added to a volume of 14 L. Thus the increase in bloodglucose, if it were all absorbed and none of it were metabolized, would be

1.PS2.3

35 g / 14 L = 2.5 g L-1 which is 2.5 g L-1 x 0.1 L dL-1 = 0.25 g dL-1 = 250 mg%

B. Given that post-prandial (after eating) increases in blood glucose amount to maybe 40 mgdL-1, depending on the meal, over a period of an hour, give a crude estimate of the rate ofglucose uptake by the peripheral tissues. Assume that the meal contains 100 g ofcarbohydrates and all of it is absorbed in one hour.

If the blood glucose increases only 40 mg%, then the amount of glucose that wasn’tabsorbed would be 40 mg dL-1 x 10 dL L-1 x 14 L = 5600 mg = 5.6 g. So the amount ofglucose which would be taken up by peripheral tissues would be 100 g - 5.6 g = 94.4 g in 1 hour.

8. The association reaction for Ca2+ and EGTA (a chemical that binds Ca2+) is written as

Ca2+ + EGTA W CaCEGTA

Under defined and particular conditions of temperature and ionic mixture, the association constantwas determined to be KA = 2.52 x 106 M-1. In a chemical mixture, 400 ìM EGTA was included and the free [Ca2+] determined by a Ca2+-selective electrode was found to be 4 x 10-7 M. Assumingthat there are no other binding agents for Ca2+, what is the total [Ca2+] in the mixture?

The associaton constant for the reaction is given as

KA = [CaCEGTA] / [Ca2+] [EGTA] = 2.52 x 106 M-1

In this case we are given that [Ca2+] = 4 x 10-7 M and that the total [EGTA] tot = 400 x 10-6M. Thequestion asks for the total [Ca2+] in the mixture, which consists of the free [Ca2+] plus [CaCEGTA]. Therefore, we need to solve for [CaCEGTA]. We let [CaCEGTA] = X ; then the free [EGTA] atequilibrium is 400 x 10-6 - X . We write

2.52 x 106 M-1 = X / 4 x 10-7 M x (4 x 10-4 - X)_

10.08 x 10-1 (4 x 10-4 - X) = X

4 x 10-4 - 1.008 X = X

4 x 10-4 = (1 + 1.008) X X = 4 x 10-4 / 2.008 = 1.992 x 10-4 M

So the total [Ca] = 1.992 x 10-4 M + 4 x 10-7M = 1.996 x 10-4M

9. 2,4-dinotrophenyl acetate decomposes in alkaline solution with a pseudo-first order rate constant of11.7 s-1 at 25E. It is a “pseudo” first order rate constant because it depends on the pH.

A. If the initial concentration of DNPA is 1 mM, what is its concentration after 15 s?

The decay of DNPA obeys the reaction

d[DNPA]/dt = - 11.7 s-1 [DNPA]

This equation can be solved by separation of variables and integration:

1.PS2.4

I d[DNPA] /[DNPA] = I - 11.7 s-1 dt

ln [DNPA] = -11.7 s-1 t + C

Taking the exponent (to e) of both sides gives

[DNPA] = [DNPA]0 e-11.7 t

Here the constant of integration is evaluated by the boundary conditions, which give that att=0 [DNPA] = [DNPA]0

We can evaluate [DNPA] at 15 s simply by inserting [DNPA]0 = 1 x 10-3 M and t= 15 s. Wefind

[DNPA] = 1 x 10-3 M e -11.7 x 15 = 6.04 x 10-80 M

This value for the concentration, although a true solution to the equation, is absurd. The concentration is actually a discrete variable, on the scale of 1 L, in increments of1.6 x 10-24 M (1/N0). This concentration is effectively zero.

B. At what time is the concentration reduced to 0.5 mM (that is, what is the half-life of thereaction)?

The half life of the reaction is the time at which [DNPA] = 1/2 [DNPA]0. This occurs when

1/2 [DNPA]0 = [DNPA]0 e-11.7 t1/2

Here the [DNPA]0 cancels; taking the ln of both sides we have

-ln 2 = -11.7 s-1 t1/2

t1/2 = ln 2 / 11.7 s-1 = 0.0592 s or 59 ms.

C. After 5 min of reaction, what is the concentration of DNPA.

We don’t even have to do this calculation to realize the concentration is zero. Theconcentration here, as a solution to the equation, will be physically meaningless.

1.PS2.5

10. The following data were obtained for the rate of the Mg-Ca-ATPase activity of vesicles of cardiacsarcoplasmic reticulum as a function of temperature. What can you tell about the activation energy?

Temperature(EC)

ATPase rate(ìmol min-1mg-1)

6.9 0.068

11.5 0.138

15.8 0.300

19.8 0.568

20.2 0.585

25.6 1.236

26.1 1.154

31.0 2.238

34.8 3.030

39.2 4.220

The Arrhenius plot is obtained by plotting the ln of the ATPase activity against the reciprocal of the absolutetemperature. The results is shown below.

Fig. 1.PS2.PROBLEM 10 ANSWER. The ln of ATPaseactivity is plotted against the inverse of the absolutetemperature. This is the Arrhenius Plot.

The slope of the Arrhenius Plot is - Ea/ R = -11388 EK Since R = 8.319 j mol-1 EK-1 , we calculate that Ea =94.74 kj mol-1, which is also 22.63 kcal mol-1. However, the graph is not a single line. It really is composedof two lines, or perhaps is a curve. The origin of this curve is not definite. It may be due to prior equilibriumor to a shift in the rate-limiting step of the enzyme with temperature.

1.PS2.6

11. Superoxide reduces cytochrome C in the reaction

Cyt CCFe3+ + O2- Y Cyt CCFe2+ + O2

Where Cyt CCFe3+ is the oxidized form and Cyt CCFe2+ is the reduced form of cytochrome C. Thereaction can be followed spectrophotometrically at 550 nm. The extinction coefficient for the reducedform of Cytochrome C is åRED = 2.99 x 104 M-1 and the extinction coefficient for the oxidized form åOX

= 0.89 x 104 M-1 (Massey, V. The microestimation of succinate and the extinction coefficient ofcytochrome C. Biochim. Biophys. Acta 34:255-256 (1959)). See 1.PS2 problem #2 for a discussionof extinction coefficients and spectrophotometry.

When xanthine oxidase converts xanthine to uric acid, it produces superoxide that can be measuredusing cytochrome C reduction. The following data were obtained for A550:

Time (min) A550

0 0.1326

1 0.1478

2 0.1637

3 0.1791

4 0.1941

5 0.2073

6 0.2202

A. Calculate the rate of cytochrome C reduction.

The linear regression of the line gives a slope of 0.01472 Absorbance units per minute. Weconvert this to mole L-1 min-1 by dividing by the difference in extinction coefficient between theoxidized and reduced form. Thus åRED - åOX = 2.99 x 104 M -1 - 0.89 x 104 M -1 = 2.10x 104 M -1 and the rate is 0.01472 min-1 / 2.10 x 104 M -1 = 7.01 x 10-7 mol L-1 min-1 which is equivalent to 0.701 x 10-9 mol mL-1 min-1

B. The xanthine oxidase was added in 75ìL of 6.5 mg XO per mL into a 3 mL reaction mixture.Calculate the specific activity of cytochrome C reduction (mols of cytochrome C reduced permin per mg of XO protein.)

The concentration of xanthine oxidase is 0.075 mL x 6.5 mg mL-1 / 3 mL = 0.1625 mg mL-1. We get the specific activity by dividing the rate of cyt C reduction by theconcentration of XO:

0.701 x 10-9 mol mL-1 min-1 / 0.1625 mg mL-1 = 4.31 x 10-9 mol min-1 mg-1

1.PS2.7

12. You suspect you are anemic and your physician orders some tests. He finds that your hemoglobinis 13g %. The molecular weight of hemoglobin is 66,500 g mol-1.

A. What is the concentration of hemoglobin in molar in your blood?

The [Hb] is 13 g% which is 13 g dL-1 x 10 dL L-1 = 130 g L-1. The molecular weight of Hbis 66,500 g mol-1 so the [Hb] in molar is

[Hb] = 130 g L-1 / 66,500 g mol-1 = 1.95 x 10-3 M

B. Each hemoglobin binds 4 oxygen molecules. If the hemoglobin is saturated with oxygen,what is the concentration of O2 bound to Hb, in molar?

The oxygen concentration ought to be 4 times the [Hb] since each Hb molecule binds 4oxygens. Thus it would be 7.82 x 10-3 M

C. Convert the answer in B to volume using the ideal gas equation, PV = nRT where T is theabsolute temperature, R = 0.082 L atm mol-1 EK-1, V is the volume that we seek and P = 1atm. The conditions for volume of gas are usually STPD - standard temperature andpressure, dry. The standard temperature is 0 EC and pressure is 1 atm.

We can convert this to units of volume %. One dL of blood would contain 7.82 x 10-4 mol(part C x 1L /10 dL). So PV = nRT where P = 1 atm, V is the unknown, n = 7.82 x 10-4

mole dL-1, R = 0.082 L atm mol-1 EK-1 and T = 273.1 EK. The volume per cent of gas at STPDis thus 0.082 L atm mol-1 EK-1 x 273.1 EK x 7.82 x 10-4 mol dL-1 = 175.1 x 10-4 L dL-1 = 17.51 mL dL-1

1.PS2.8

13. The rate of ATP hydrolysis by ATPases can be followed by the coupled enzyme assay shown below:

Fig. 1.PS2.2 ATP hydrolysis of pyruvate kinase convertsphosphoenolpyruvate to pyruvic acid. This is coupled by lacticdehydrogenase to the conversion of pyruvic acid to lactic acidand conversion of NADH to NAD+. The progress of thereaction can be followed spectrophotometrically by the changein absorbance of NADH.

The progress of the reaction can be followed by A340. The extinction coefficient of NAD+ at 340 nmis negligible. The exctinction coeffcient of NADH at 340 nm is 6.2 x 103 M-1. See 1.PS2.2 problem#2 for a discussion of extinction coefficients and spectrophotometry. In one reaction, theconcentration of Ca-ATPase was 0.22 mg mL-1 and A340 was 0.65 at t=0 min and 0.455 at t=2.0 min. What is the activity of the Ca-ATPase in units of ìmol min-1 mg-1?

The decrease in [NADH] is given as ( 0.65 - 0.455)/ 6.2 x 103 M-1 = 0.0315 x 10-3 M = 0.0315 x 10-6 mol mL-1, but this decrease occurs in 2 min. So the rate of decrease is 0.0157 x 10-6

mol mL-1 min-1. The Ca-ATPase activity is obtained by dividing this rate of decrease in NADH by theamount of protein, which was 0.22 mg mL-1. So we have

0.0157 x 10-6 mol min-1 mL-1 / 0.22 mg mL-1 = 0.0715 x 10-6 mol min-1 mg-1

1.PS2.9

14. Show by representative calculations that Stirling’s formula

is a good approximation for n!. Use n=1,2,3,4,5

Calculations of n! And p2ðn enln - n are given in Table 1.PS2.2 below:

n n! Stirling’s Formula

1 1 0.92

2 2 1.92

3 6 5.84

4 24 23.51

5 120 118.02

6 720 710.08

7 5040 4980.4

8 40320 39902.4

9 362880 359536

10 3628800 3598694

15 Show that the equation

obeys Fick’s Second Law of Diffusion.

This equation can be shown to obey Fick’s Second Law of Diffusion by taking M /dt and M2/Mx2 andshowing their equivalence. When taking the derivatives, we carefully apply the chain rule:

1.PS2.10

A comparison of the last lines in these two derivatives gives us Fick’s Second Law of Diffusion:

16. The intestinal enterocytes form a covering over the intestinal lining which, to the first approximation,can be considered to be a plane. Assuming no binding or sequestration within the cell, what is theestimated time of diffusion of Ca2+ across the intestinal enterocyte? The length of the enterocyte is20 ìm and assume that the effective diffusion coefficient of Ca2+ is about 0.4 x 10-5 cm2s-1.

The time for one-dimensional diffusion is related to the distance by

where x is the distance, D is the diffusion coefficient and t is the time of diffusion. In the problem, D= 0.4 x 10-5 cm2s-1 and x = 20 x 10-4 cm. Inserting these values, we calculate

t = (20 x 10-4 cm)2 / ( 2 x 0.4 x 10-5 cm2s-1) = 0.5s

1.PS2.11

17. Table 1.PS2.1 lists the diffusion coefficients and the molecular weight of a variety of proteins. Whatrelationship can you deduce between the size and the diffusion coefficients of these soluble proteins? (Hint: regress ln D against ln Mr.). Is the relationship you found consistent with the Stokes-Einsteinequation?

Table 1.PS2.1 Diffusion coefficients and Mr for a variety of proteins

Protein Molecular Weight D x 107 (cm2s-1)

milk lipase 6,600 14.5

Metallothionein 9,700 12.4

Cytochrome C 12,000 12.9

Ribonuclease 12,600 13.1

Myoglobin 16,890 11.3

Chymotrypsinogen 23,200 9.5

Carbonic anhydrase 30,600 10.0

Peroxidase II 44,050 6.8

Albumin 68,500 6.1

Lactoperoxidase 92,620 6.0

Aldolase 149,100 4.6

The Stokes-Einstein equation states that the diffusion coefficient is given by

where D is the diffusion coefficient, k is Boltzmann’s constant, T is the absolute temperature, ð is thegeometric ratio, ç is the viscosity and a is the radius of the molecule, assuming it is a sphere. As themolecular size increases, a increases and D should decrease. All other variables are unchangedamong the different proteins. What we are searching for is the relationship between D and a. If themolecules are spherical, and their densities are constant, then molecular weight should beproportional to the volume occupied. Thus we have

where Mr is the molecular weight (in g mol-1), N0 is Avogadro’s number (molecules mol-1) and v is thepartial specific volume equal to the inverse of the density, in units of cm3 g-1. According to thisrelationship, we could re-write a as

We can insert this back into the Stokes=Einstein equation to get

1.PS2.12

All of the terms to the left of the Mr term are constants. Thus we would expect the diffusion coefficient to beinversely proportional to the cube root of the molecular weight. We can determine the empirical relationshipby plotting ln D against ln Mr. In this case we expect

The plot of ln D against ln Mr for the data supplied is shown at theleft. The slope of the line is -0.379. Thus the results are not anexact fit to the Stokes-Einstein relation, but they seem to be areasonably good fit. This may be in part because these proteinsare not spherical.

Fig. 1.PS2. PROBLEM 17 ANSWER. As suggested by the Stokes-Einstein equation, the natural log of the diffusion coefficient was plottedagainst the natural log of the molecular weight.

18. The free diffusion coefficient of oxygen in aqueous solutions is about 1.5 x 10-5 cm2 s-1. If the diffusiondistance between air and blood is 0.5 ìm, about how long is the diffusion time?

The time is estimated from Ät = x2 / 2 D = ( 0.5 x 10-4 cm)2 / 2 x 1.5 x 10-5 cm2s-1 =

0.25 x 10-8 cm2 / 3 x 10-5 cm2 s-1 = 0.083 x 10-3 s = 0.083 ms

19. Suppose a soluble protein has a molecular weight of 45 kDa and a density of 1.06 g cm-3. Supposefurther that the viscosity of the cytoplasm has a viscosity of 0.005 Pa s (about five times that of water -there is debate about the viscosity of cytoplasm with numbers varying from 0.001 to over .1 Pa s).

A. Estimate the diffusion coefficient for the protein in the cytoplasm at 37EC

Here we use the Stokes-Einstein equation: D = kT / 6ðçr We are given the viscosity (ç = 0.05Pa s) and the temperature (T = 310 oK) and k = 1.38 x 10-23 joule oK-1 ; we estimate r from the density

1.PS2.13

and the molecular weight. The molar volume = No V = No 4/3 ðr3 = 45,000 g / 1.06 g cm-3 = 42543 cm3; we calculate r to ber = 25.6 x 10-8cm = 2.56 x 10-9 m .

Plugging this into the Stokes-Einstein equation, we have

D = 1.38 x 10-23 joules oK-1 x 310 oK / 6ð x 0.005 N m-2 s x 2.56 x 10-9 m = 1.77 x 10-11 m2s-1

= 1.77 x 10-7 cm2 s-1

B. If the protein were synthesized in the cell body, or soma, of a neuron in the spinal cord, abouthow long would it take to diffuse to the axon terminal 75 cm away?

Here t = x2 / 2D = ( 75 cm)2 / 1.77 x 10-7 cm2 s-1 = 3.18 x 1010 s = 5.3 x 108 min

20. Diffusion coefficients in cytoplasm has been estimated by a technique of photobleaching recovery. In this technique, an area of the cytoplasm is irradiated with light to photobleach a fluorescent probe. Recovery of fluorescence in the region is achieved by diffusion of unbleached probes from adjacentareas of the cytoplasm. The translational diffusion coefficient can be estimated from the half-time offluorescent recovery. (Axelrod, D., et al., Mobility measurements by analysis of fluorescencephotobleaching recovery kinetics. Biophys. J. 16:1055-1069 (1976)). This technique was applied toestimate the relative viscosity of cytoplasm and nucleoplasm by microinjecting fluoresceinisothiocyanate-labeled dextrans of varying molecular sizes and measuring the fluorescencephotobleaching recovery (Lang, I., et al., Molecular mobility and nucleoplasmic flux in hepatoma cells.J. Cell Biol. 102:1183-1190 (1986)). These authors obtained the following data:

Probe MolecularWeight

Equivalentradius

D in Dilutesolution

D in cytoplasm D in nucleoplasm

(kD) (nm) D is in units of 10-6 cm2s-1

FD20 17.5 3.30 0.651 0.080 ---

FD40 41.0 4.64 0.463 0.044 0.069

FD70 62.0 5.51 0.390 0.029 0.056

FD150 156.9 9.07 0.237 0.015 0.036

A. Plot D against 1/a, where a is the molecular radius, for each of the solutions. From theStokes-Einstein relation, you would expect the resulting curves to pass through the origin ofzero diffusion coefficient with infinite radius. Do the curves extrapolate back in this way? Why or why not?

1.PS2.14

Fig. 1.PS2. PROBLEM 20 ANSWER. Theslope of the curve of D against 1/a, where a isthe Stoke’s radius, is linear for all threesituations: solution, cytoplasm andnucleoplasm. The curve extrapolates back tozero for the solution and nucleoplasm, and notquite to zero for the cytoplasm.

The curves for the diffusion coefficient determined in solution extrapolate to the origin, whichcorresponds to a zero diffusion coefficient for an infinitely large solute. This is also the case for thenucleoplasm, but is NOT the case for the cytoplasm. The error, however, is not large, and confidence in theresults is not high. A possible cause for deviation from the origin intercept is disobedience to the Stokes-Einstein equation. This could be due to diffusional obstacles in the cytoplasm that artificially lower thediffusion coefficient for larger solutes.

B. Regardless of the intercept, the slope of the plot from part A ought to be related to theviscosity of the medium. Use the slopes to estimate the relative viscosity of the dilutesolution, cytoplasm and nucleoplasm.

According to the Stokes-Einstein equation,

D = kT / 6ðça the slope of the line should be b(1) = kT/ 6ðç . Therefore we can calculate

ç = kT/ 6ð b(1)

Using k = 8.314 joules mol-1 EK-1 / 6.02 x 1023 mol-1 = 1.381 x 10-23 joules EK-1, T = 310EK, we calculatethat

ç = 2.271 x 10-22 joule / b(1)

The slope of the plots given in part A are:

solution: b(1) = 2.148 x 10-6 cm2s-1 nm = 2.148 x 10-6 cm2 s-1 nm x 10-4 m2 cm-2 x 10-9 m nm-1

= 2.148 x 10-19 m3 s-1

Inserting this into the equation for ç = 2.271 x 10-22 joule / 2.148 x 10-19 m3 s-1 gives

ç = 1.05 x 10-3 N m-2 s = 0.00105 Pa s for the solution

This is close to that of water, 8.9 x 10-4 Pa s.

1.PS2.15

For the cytoplasm, the slope is 3.438 x 10-7 cm2 s-1 nm x 10-4 m2 cm-2 x 10-9 m nm-1

= 3.438 x 10-20 m3 s-1

Solving for the viscosity, we obtain ç = 2.271 x 10-22 joule / 3.438 x 10-20 m3 s-1 giving

ç = 0.661 x 10-2 N m-2 s = 0.00661 Pa s for the cytoplasm

which is about 5 times that of water.

For the nucleoplasm, the slope is 3.085 x 10-7 cm2 s-1 nm x 10-4 m2 cm-2 x 10-9 m nm-1

= 3.085 x 10-20 m3 s-1

Solving for the viscosity, we obtain ç = 2.271 x 10-22 joule / 3.085 x 10-20 m3 s-1 giving

ç = 0.736 x 10-2 N m-2 s = 0.00736 Pa s for the nucleoplasm

which is about 7 times that of water.

1.PS2.16

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